Five-Card Straight

Date: 12/10/2002 at 13:23:01
From: David Stadler
Subject: Poker probability
From a standard 52-card deck you pull out a 5-6-7-8 (suit is
irrelevent for the problem). You shuffle the remaining 48 cards and
deal 3 more cards. What percentage of the time will you complete the
5-card straight (by receiving a 4 or 9 with any one of the three cards
dealt). My friends and I disagree on the answer. Here is my math:
1st draw = 16.6667%
2nd draw = 17.0213%
3rd draw = 17.3913%
adding the three chances together I get 51.08% you will get a 4 or 9.
My reasoning:
Ignoring the slight increase in percentage when a non-winning card is
dealt, your odds are 8/48 or 1/6, similar to dice. If I select one
number from a die and roll 3 times (3 cards), my odds are 50% to roll
my number.
Please help us settle this dispute - it is amazing how stubborn we all
are!

Date: 12/10/2002 at 20:20:07
From: Doctor Ian
Subject: Re: Poker probability
Hi David,
The problem with these 'at least one' situations is that you have to
consider _all_ of the possibilities:
p(4,-,-) '-' = not 4 or 9
+ p(-,4,-)
+ p(-,-,4)
+ p(9,-,-)
+ p(-,9,-)
+ p(-,-,9)
+ p(4,9,-)
+ p(9,4,-)
+ ...
It's a headache, to say the least. Fortunately, there is a trick you
can use that works very nicely:
p(at least once) = 1 - p(never)
Does that make sense?
There are four 4's and four 9's remaining in the deck, which has 48
cards. So the probability of drawing something other than a 4 or a 9
would be
(48 - 8)/48
The probability of doing it again would be
(47 - 8)/47
And again,
(46 - 8)/46
So the probability of missing on all three draws would be
40 39 38
-- * -- * --
48 47 46
5 39 19
= -- * -- * --
6 47 23
5 13 19
= -- * -- * --
2 47 23
= 1235/2162
= 0.57
So the probability of hitting at least one would be
p = 1 - 0.57
= 0.43
Does this help?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

Date: 12/12/2002 at 10:16:39
From: David Stadler
Subject: Thank you (Poker probability)
Dr. Ian -
Thank you very much for explaining the poker probability question to
me. I made a program to simulate and came up with the same answer
before I sent the question, but still couldn't see what I was missing.
Thanks again.
D. Stadler